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2 simple derivative problems

  • #1

Homework Statement


Find the derivative of the function
F(z) = [tex]\sqrt{\frac{z-1}{z+1}}[/tex]

Homework Equations






The Attempt at a Solution



[tex]\frac{z-1}{z+1}[/tex][tex]^{1/2}[/tex]

1/2 [tex]\frac{z-1}{z+1}[/tex][tex]^{-1/2}[/tex]

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement


Find the derivative of the function.
y = tan(sin2x)

Homework Equations





The Attempt at a Solution



[tex]\frac{sin}{cos}[/tex]

I'm assuming this is the chain rule, but I'm not sure how to approach it.

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
You've also forgotten to use the chain rule on the first problem. (f(g(x))'=f'(g(x))*g'(x). You forgot the g'(x). Once you've straightened that out try the second.
 
  • #3
soooo 1/2 [tex]\frac{z-1}{z+1}[/tex][tex]^{-1/2}[/tex] (1) ?
 
  • #4
be careful with your brackets. the square root is being applied to (z-1/z+1).
This is a chain rule question, what is your outside function? your inside function?
 
  • #5
Gib Z
Homework Helper
3,346
4
It might help to call the things inside the square root, [tex]\frac{z-1}{z+1}[/tex] u. That way the chain rule basically says that the derivative of the whole thing is the derivative of F(z) with respect to u* derivative of u with respect to z.
 
  • #6
danago
Gold Member
1,122
4
For a question like the first one, i tend to find it less tedious to first take natural logs of both sides, and then using log laws, reduce it down to a form that is easier to work with, and then differentiate implicitly. Thats if you have learnt implicit differentiation though.
 
  • #7
In the second question, I'm not sure you need to use the identity.

[tex]\frac{sin}{cos}=tan[/itex]

It seems just a simple derivitive:-

[tex]\frac{d}{dx}tan=sec^2[/itex]

Would be fine here?
 
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  • #8
VietDao29
Homework Helper
1,423
1
soooo 1/2 [tex]\frac{z-1}{z+1}[/tex][tex]^{-1/2}[/tex] (1) ?
No, that's not correct.

Why is it (1)? As others have pointed out, you should use The Chain Rule:

[tex]\frac{df}{dx} = \frac{df}{du} \times \frac{du}{dx}[/tex], i.e the derivative of f with respect to x is the derivative of f with respect to u times the derivative of u with respect to x. (u is any function that you choose, so that the problem becomes easier). There are problems that you have to apply the rules twice, threefold, or even more:

[tex]\frac{df}{dx} = \frac{df}{du_1} \times \frac{du_1}{dx} = \frac{df}{du_1} \times \left( \frac{du_1}{du_2} \times \frac{du_2}{dx} \right) = \frac{df}{du_1} \times \frac{du_1}{du_2} \times \left( \frac{du_2}{du_3} \times \frac{du_3}{dx} \right) = ...[/tex]

Example:
Find the derivative of [tex]\tan \left( \sqrt{\cos x} \right)[/tex] with respect to x.

-------------------

Now that looks monstrous. But if you let [tex]u = \sqrt{\cos x}[/tex], the whole thing becomes: tan(u).

Apply The Chain Rule here, we have:
[tex]\tan \left( \sqrt{\cos x} \right)'_x = \tan \left( u \right)'_x = \tan \left( u \right)'_u \times u'_x = \frac{1}{\cos ^ 2 u} \times (\sqrt{\cos x})'_x[/tex]

Apply the Rule once more, let t = cos(x)
[tex]... = \frac{1}{\cos ^ 2 u} \times (\sqrt{\cos x})'_x = \frac{1}{\cos ^ 2 u} \times (\sqrt{t})'_x = \frac{1}{\cos ^ 2 u} \times (\sqrt{t})'_t \times t'_x = \frac{1}{\cos ^ 2 u} \frac{1}{2 \sqrt{t}} (\cos x)'_x = -\frac{1}{\cos ^ 2 u} \frac{1}{2 \sqrt{t}} \sin x[/tex]

Now, change u, and t back to x's, we have:
[tex]... = -\frac{1}{\cos ^ 2 u} \frac{1}{2 \sqrt{t}} \sin x = -\frac{1}{\cos ^ 2 (\sqrt{\cos x})} \times \frac{1}{2 \sqrt{\cos x}} \sin x[/tex].

To make it look a little bit nicer. You can pull the 1/2 factor in front of the whole expression, like this:

[tex]... = - \frac{1}{2} \times\frac{1}{\cos ^ 2 (\sqrt{\cos x})} \times \frac{\sin x}{\sqrt{\cos x}}[/tex]


It's done. :)

Can you do the same to your 2 problems? :)
 
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  • #9
28
0
for the first problem u could use logs to differentiate such that lny = 1/2ln(z-1) - 1/2ln(z+1) and then implicitly differentiate, dy/dz.1/y = 1/2.(1/z-1) - 1/2(1/z+1). dy/dz=[1/(2z-2)-1/(2z+2)].y = [4/(4z^2+4)].root[(z-1)/(z+1)] and simplify.

u = sin2x therefore y=tan (u) therefore dy/du = u'/cos^2 (u) = 2cos(2x)/cos^2(sin2x)
 
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